is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.


Simple Definition

This is a non-mathematical, non-academic definition of continuous: you can draw your function without lifting your pencil.



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Actual Mathematical Definition

MathJax TeX Test Page Recall that this is the definition of a limit: $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. if } 0 < ||x-c|| < \delta \to ||f(x) - f(c)|| < \epsilon$$ You might see it written out differently, but the definition of continuous is very similar $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. if } ||x-c|| < \delta \to ||f(x) - f(c)|| < \epsilon$$ The only difference is the $0 <$ part. Why is that? Think about the first picture. If the x is super close (but not equal) to $x_0$, the f(x) values will approach the red point. It will work as expected.

Now, think about the continuous definition. If we consider a super small $\epsilon$, then if $x = x_0$, which less than any $\delta > 0$, the expected f(x) values and the jump point will be greater than epsilon.

Not Continuous

MathJax TeX Test Page Let's say we want to prove that something isn't continuous. We start with the definition of continuous: $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. if } ||x-c|| < \delta \to ||f(x) - f(c)|| < \epsilon$$ Now, think about the opposite. We have to show that this definition doesn't hold for at least one epsilon. $$\exists \epsilon > 0 \dots$$ Now what? Well, we want to show that there doesn't exist a $\delta$ that satisfies the continuous definition. $$\exists \epsilon > 0, \forall \delta > 0 \text{ s.t. }\exists x \text{ s.t. } ||x-c|| < \delta \text{ and } ||f(x) - f(c)|| \ge \epsilon$$


MathJax TeX Test Page Prove the heaviside function is not continuous at x = 0. $$f(x) = \begin{cases}1 & x \ge 0 \\0 & x \lt 0\end{cases}$$ Now, we pick an $\epsilon$. We can show this doesn't work for any $\epsilon < 1$. Let's say $\frac{1}{2}$. $$\text{Let } \epsilon = \frac{1}{2}$$ Now, we have to prove it for all $\delta > 0$, so we say $$\text{Let } \delta > 0 $$ We now have a ball around 0. We want to show that a point x is within that ball, but the difference between that and $f(0) > \frac{1}{2}$. So, we have to pick a negative x value. $$\text{Let } x = -\frac{\delta}{2}$$ $$||x - 0|| = \frac{\delta}{2} < \delta$$ $$||f(x) - f(0)|| = ||0 - 1|| = 1 > \dfrac{\epsilon}{2}$$ Now, we know there exists an $\epsilon > 0$ ($\frac{1}{2}$) such that for every $\delta > 0$, there exists an x that is within $\delta$ of 0 and f(x) is not within $\epsilon$ of $f(0)$.

A Function is Continuous

That means that the function is continuous at every point in its domain.

Differentiability Implies Continuity

We will prove this statement in two steps. First, we show the Lipschitz Condition is true. Now, once we show that, we can show that it’s continuous.

Lipschitz Condition

MathJax TeX Test Page The Lipschitz condition states $$\exists C > 0 \text{ and } \delta > 0 \text{ s.t. if } ||h|| < \delta, ||f(x + h) - f(x)|| < C||h||$$ Proof:

We know that f(x) is differentiable. That's a given. So, we begin with this $$\lim_{h \to 0} \frac{||f(p + h) - f(p) - Df(p)(h)||}{||h||} = 0$$ Because this is a limit, we can pick a $\delta$ such that the above fraction will be less than 1. $$\frac{||f(p + h) - f(p)||}{||h||} \leq \frac{||f(p + h) - f(p) - Df(p)(h)||}{||h||} + \frac{||Df(p)(h)||}{||h||} $$ (by the Triangle Inequality Theorem). $$\frac{||f(p + h) - f(p)||}{||h||} \lt 1 + \frac{||Df(p)(h)||}{||h||} $$ In the post about total derivatives, we prove not only that the derivative is a linear transformation but also that this inequality holds:$$\exists C>0 \text{ s.t. } ||T(x)|| \leq C||x||$$ We now write this $$\frac{||f(p + h) - f(p)||}{||h||} \lt 1 + \frac{C_1||h||}{||h||} $$ $$\frac{||f(p + h) - f(p)||}{||h||} \lt 1 + C_1$$ Now, we let C = $1 + C_1$ $$||f(p + h) - f(p)|| \leq \left(1 + C\right)||h||$$ Therefore, the Lipschitz Condition is satisfied.

Proving Continuity from Lipschitz

MathJax TeX Test Page $$\text{Let }\epsilon > 0$$ $$\text{Let }\delta = \frac{\epsilon}{C}$$ $$\text{Assume } ||x - p||<\delta$$ $$||x - p||<\frac{\epsilon}{C}$$ $$C||x-p|| < \epsilon$$ Now, we let $h = x - p$. $$||f(p + h) - f(p)|| < C||h|| < \epsilon$$ We substitute $x-p$ back in. $$||f(x) - f(p)|| < \epsilon$$ We can now combine this entire proof. $$\forall \epsilon > 0 \exists \delta \text{ s.t. if } ||x-p|| < \delta \to ||f(x) - f(p)|| < \epsilon$$ That is the definition of continuous.

Lipschitz Continuity

In fact, if a function satisfies the Lipschitz condition, it is considered Lipschitz continuous. An easy way to see if something is Lipschitz continuous is if the derivative is bounded. Why is that? Well the Lipschitz condition states that you can create a line that is greater than or equal to the function. If the derivative is unbounded, then the function will exceed that line.

MathJax TeX Test Page $$f(x) = \sqrt{x} \to f'(x) = \dfrac{1}{2\sqrt{x}}$$ In this case, as $x \to 0$, $f'(x) \to \infty$. There is no constant that is greater than that, so it's not Lipschitz. $$f(x) = x^2 \to f'(x) = 2x$$ As $x \to \infty$, $f'(x) \to \infty$. No constant can be made greater than that, so it's not Lipschitz. $$f(x) = \dfrac{1}{\sqrt{x^2 + 5}} \to f'(x) = \dfrac{x}{\sqrt{x^2 + 5}}$$ That is bounded by -1 and 1.
David Witten

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